1. **Stating the problem:** We are given the function $$V(n) = c (1 - 2N n)^n$$ and asked to find its derivative $$V'(n)$$ and then compute $$\frac{V'(n)}{V(n)}$$. 2. **Formula and rules:** To differentiate $$V(n) = c (1 - 2N n)^n$$ where both the base and the exponent depend on $$n$$, we use logarithmic differentiation. 3. **Step 1: Take the natural logarithm of both sides:** $$\ln V(n) = \ln c + n \ln(1 - 2N n)$$ 4. **Step 2: Differentiate both sides with respect to $$n$$:** $$\frac{V'(n)}{V(n)} = 0 + \ln(1 - 2N n) + n \cdot \frac{1}{1 - 2N n} \cdot (-2N)$$ 5. **Step 3: Simplify the derivative:** $$\frac{V'(n)}{V(n)} = \ln(1 - 2N n) - \frac{2N n}{1 - 2N n}$$ 6. **Step 4: Express $$V'(n)$$ explicitly:** $$V'(n) = V(n) \left( \ln(1 - 2N n) - \frac{2N n}{1 - 2N n} \right) = c (1 - 2N n)^n \left( \ln(1 - 2N n) - \frac{2N n}{1 - 2N n} \right)$$ **Summary:** - The derivative is $$V'(n) = c (1 - 2N n)^n \left( \ln(1 - 2N n) - \frac{2N n}{1 - 2N n} \right)$$. - The ratio is $$\frac{V'(n)}{V(n)} = \ln(1 - 2N n) - \frac{2N n}{1 - 2N n}$$. This method uses logarithmic differentiation to handle the variable exponent and base.